The manufacture of aluminum includes the production of cryolite (Na3AlF6) from the following reaction:

6HF+3NaAlO2=Na3AlF6+3H2O+Al2O3

How much NaAlO2 (sodium aluminate) is required to produce 3.47 kg of Na3AlF6?

6HF + 3 NaAlO2 = Na3AlF6 + 3H2O + Al203

calculate moles of Na3AlF6 formed

moles = mass/molar mass

molar mass of Na3AlF6 = 209.94 g/mol

convert 3.47Kg to grams, that is 3.47 x1000 = 3470 grams

moles is therefore = 3470 g / 209.94 g/mol = 16.53 moles

by use of mole ratio between NaAlO2 to Na3AlF6 which is 3:1 therefore the moles of NaAlO2 is (16.53 x3) / 1 = 49.59 moles

mass of NaAlO2 is = moles of NaAlO2 x molar mass NaAlO2

that is 49.59 mole x 81.97 g/mol = 4064.89 grams or 4.065 Kg